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304. Range Sum Query 2D - Immutable

Description

Given a 2D matrix matrix, handle multiple queries of the following type:

  • Calculate the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

Implement the NumMatrix class:

  • NumMatrix(int[][] matrix) Initializes the object with the integer matrix matrix.
  • int sumRegion(int row1, int col1, int row2, int col2) Returns the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

You must design an algorithm where sumRegion works on O(1) time complexity.

 

Example 1:

Input
["NumMatrix", "sumRegion", "sumRegion", "sumRegion"]
[[[[3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3, 0, 5]]], [2, 1, 4, 3], [1, 1, 2, 2], [1, 2, 2, 4]]
Output
[null, 8, 11, 12]

Explanation
NumMatrix numMatrix = new NumMatrix([[3, 0, 1, 4, 2], [5, 6, 3, 2, 1], [1, 2, 0, 1, 5], [4, 1, 0, 1, 7], [1, 0, 3, 0, 5]]);
numMatrix.sumRegion(2, 1, 4, 3); // return 8 (i.e sum of the red rectangle)
numMatrix.sumRegion(1, 1, 2, 2); // return 11 (i.e sum of the green rectangle)
numMatrix.sumRegion(1, 2, 2, 4); // return 12 (i.e sum of the blue rectangle)

 

Constraints:

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 200
  • -104 <= matrix[i][j] <= 104
  • 0 <= row1 <= row2 < m
  • 0 <= col1 <= col2 < n
  • At most 104 calls will be made to sumRegion.

Solutions

Solution 1: Two-dimensional Prefix Sum

We use $s[i + 1][j + 1]$ to represent the sum of all elements in the upper left part of the $i$th row and $j$th column, where indices $i$ and $j$ both start from $0$. We can get the following prefix sum formula:

$$ s[i + 1][j + 1] = s[i + 1][j] + s[i][j + 1] - s[i][j] + nums[i][j] $$

Then, the sum of the elements of the rectangle with $(x_1, y_1)$ and $(x_2, y_2)$ as the upper left corner and lower right corner respectively is:

$$ s[x_2 + 1][y_2 + 1] - s[x_2 + 1][y_1] - s[x_1][y_2 + 1] + s[x_1][y_1] $$

In the initialization method, we preprocess the prefix sum array $s$, and in the query method, we directly return the result of the above formula.

The time complexity for initializing is $O(m \times n)$, and the time complexity for querying is $O(1)$. The space complexity is $O(m \times n)$.

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class NumMatrix:
    def __init__(self, matrix: List[List[int]]):
        m, n = len(matrix), len(matrix[0])
        self.s = [[0] * (n + 1) for _ in range(m + 1)]
        for i, row in enumerate(matrix):
            for j, v in enumerate(row):
                self.s[i + 1][j + 1] = (
                    self.s[i][j + 1] + self.s[i + 1][j] - self.s[i][j] + v
                )

    def sumRegion(self, row1: int, col1: int, row2: int, col2: int) -> int:
        return (
            self.s[row2 + 1][col2 + 1]
            - self.s[row2 + 1][col1]
            - self.s[row1][col2 + 1]
            + self.s[row1][col1]
        )


# Your NumMatrix object will be instantiated and called as such:
# obj = NumMatrix(matrix)
# param_1 = obj.sumRegion(row1,col1,row2,col2)

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class NumMatrix {
    private int[][] s;

    public NumMatrix(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        s = new int[m + 1][n + 1];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                s[i + 1][j + 1] = s[i + 1][j] + s[i][j + 1] - s[i][j] + matrix[i][j];
            }
        }
    }

    public int sumRegion(int row1, int col1, int row2, int col2) {
        return s[row2 + 1][col2 + 1] - s[row2 + 1][col1] - s[row1][col2 + 1] + s[row1][col1];
    }
}

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix obj = new NumMatrix(matrix);
 * int param_1 = obj.sumRegion(row1,col1,row2,col2);
 */

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class NumMatrix {
public:
    vector<vector<int>> s;

    NumMatrix(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        s.resize(m + 1, vector<int>(n + 1));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                s[i + 1][j + 1]